Timeline for Remainders $\quad 1\quad 2\quad $ only
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2013 at 6:44 | vote | accept | Włodzimierz Holsztyński | ||
| Jun 29, 2013 at 10:24 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Added GAP code used for computing the listed values.
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| Jun 27, 2013 at 7:21 | comment | added | Włodzimierz Holsztyński | @Gerry, our own MO Stefan Kohl got further than OEIS! Good reflex and association, Gerry, great links--thank you. | |
| Jun 27, 2013 at 5:54 | comment | added | Gerry Myerson | The numbers $n(p)-2$ are tabulated at oeis.org/A118478 and the ratios are tabulated at oeis.org/A215021 but neither page has anything about rate of growth. | |
| Jun 27, 2013 at 3:29 | comment | added | Yoav Kallus | Here are the ratios $[n(p)-1][n(p)-2]/P(p)$ (necessarily integers) for $p=3,5,\ldots,97$: 1, 1, 1, 19, 17, 1, 409, 604, 82, 20951, 229931, 411012, 39080794, 4382914408, 6345486566, 45119290746, 581075656330, 8672770990, 869561574799171, 71853663603175593, 25509154378676494, 24040267482771436703, 102403319155457392955, 11302410854347731819765 | |
| Jun 26, 2013 at 12:00 | comment | added | Włodzimierz Holsztyński | Let $p<q$ be consecutive odd primes. If $$\frac{n(q)}{n(p)}<\sqrt q$$ then--by definition--the jump is low; otherwise the jump is high (the inequality is always sharp). It seemed to me that low/high depends (but not consistently) on the $\mod 4$ class of $q$. When $q=3\mod 4$ then the jump is either extra low or extra high depending on the parity of the number of primes $s < q$ which are congruent to $3\mod 4$. Your, @Stefan, list seems to confirm this conjecture. There should be also other considerations which occasionally mess up this tendency; it still should be the most common tendency. | |
| Jun 26, 2013 at 11:48 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Added values n(p) for p = 79, 83, 89, 97.
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| Jun 26, 2013 at 10:14 | history | answered | Stefan Kohl♦ | CC BY-SA 3.0 |